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    Hello
    Just been going through some past papers for revision for Tuesday, and got stuck on part b of the attached question. Any help would be greatly appreciated!

    If it's any help, the correct answer to part a) is 3/2x^2 - 10x^1/2 - 2x +9

    Thanks in advance!
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    (Original post by TheGirlWhoLived)
    Hello
    Just been going through some past papers for revision for Tuesday, and got stuck on part b of the attached question. Any help would be greatly appreciated!

    If it's any help, the correct answer to part a) is 3/2x^2 - 10x^1/2 - 2x +9

    Thanks in advance!
    Tangent means it has the same gradient as the curve. You can plug the x coordinate of that point into the dy/DX given to work out the gradient of tangent then use y=MX +c

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    (Original post by samb1234)
    Tangent means it has the same gradient as the curve. You can plug the x coordinate of that point into the dy/DX given to work out the gradient of tangent then use y=MX +c

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    Thanks, I'll give it a try! Seems so simple now you've pointed it out
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    (Original post by TheGirlWhoLived)
    Thanks, I'll give it a try! Seems so simple now you've pointed it out
    Let me know if you get stuck

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    Calculate the value of dy/dx when x = 4

    This will give you the gradient of the tangent at the point (4,5)

    Now you know the gradient of the tangent and you also know a pair of points (4,5) which the tangent passes through

    Now use y-y1 = m(x-x1) to find the equation of the tangent at the point P

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